where $$C$$ is a constant. Because that the given family members of curves, we can attract the orthogonal trajectories, the is one more family of curves $$f\left( x,y \right) = C$$ that cross the given curves at right angles.

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For example, the orthogonal trajectory that the family of straight lines defined by the equation $$y = kx,$$ whereby $$k$$ is a parameter (the steep of the straight line), is any circle having facility at the origin (Figure $$1$$):

where $$R$$ is the radius the the circle. Figure 1.

Similarly, the orthogonal trajectories the the family members of ellipses

\<\fracx^2a^2 + \fracy^2c^2 - a^2 = 1,\;\; \textwhere\;\; 0 \lt a \lt c,\>
\<\fracx^2b^2 - \fracy^2b^2 - c^2 = 1,\;\; \textwhere\;\; 0 \lt c \lt b.\>

Both families of curves space sketched in number $$2.$$ right here $$a$$ and also $$b$$ play the role of parameters explicate the family members of ellipses and also hyperbolas, respectively. Figure 2.

## General technique of recognize Orthogonal Trajectories

The common technique for determining orthogonal trajectories is based on solving the partial differential equation:

where the prize $$\nabla$$ method the gradient that the function $$f\left( x,y \right)$$ or $$g\left( x,y \right)$$ and the dot way the dot product that the 2 gradient vectors.

Using the definition of gradient, one have the right to write:

\<\nabla f\left( x,y \right) = \mathbfgrad\,f\left( x,y \right) = \left( \frac\partial f\partial x,\frac\partial f\partial y \right),\>
\<\nabla g\left( x,y \right) = \mathbfgrad\,g\left( x,y \right) = \left( \frac\partial g\partial x,\frac\partial g\partial y \right).\>
\<\nabla f\left( x,y \right) \cdot \nabla g\left( x,y \right) = 0,\;\; \Rightarrow \left( \frac\partial f\partial x,\frac\partial f\partial y \right) \cdot \left( \frac\partial g\partial x,\frac\partial g\partial y \right) = 0,\;\; \Rightarrow \frac\partial f\partial x\frac\partial g\partial x + \frac\partial f\partial y\frac\partial g\partial y = 0.\>

Solving the critical PDE, we deserve to determine the equation the the orthogonal trajectories $$f\left( x,y \right) = C.$$

## A handy Algorithm for creating Orthogonal Trajectories

Below we explain an much easier algorithm because that finding orthogonal trajectories $$f\left( x,y \right) = C$$ the the given family members of curves $$g\left( x,y \right) = C$$ using just ordinary differential equations. The algorithm includes the complying with steps:

Replace $$y"$$ with $$\left( - \frac1y" \right)$$ in this differential equation. Together a result, we obtain the differential equation that the orthogonal trajectories.Solve the brand-new differential equation to determine the algebraic equation the the household of orthogonal trajectories $$f\left( x,y \right) = C.$$

## Solved Problems

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### Example 1

Find the orthogonal trajectories of the family members of straight lines $$y = Cx,$$ where $$C$$ is a parameter.

### Example 2

A family members of hyperbolic curve is provided by the equation $$y = \fracCx.$$ uncover the orthogonal trajectories because that these curves.

### Example 1.

Find the orthogonal trajectories the the household of straight lines $$y = Cx,$$ where $$C$$ is a parameter.

We apply the algorithm explained on the previous page.

$$1)$$ First, we construct the differential equation because that the household of directly lines $$y = Cx.$$ By distinguishing the last equation with respect come $$x,$$ we get:

In the critical equation we changed $$2C$$ with simply a constant $$C.$$ Thus, we have acquired the equation that the family of orthogonal trajectories. Together it deserve to be seen, this orthogonal trajectories are likewise hyperbolas. Both the families of hyperbolas are presented schematically in figure $$3.$$